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3 years ago

Write a polynomial in standard form that represents the area of the shaded region

Mathematics

2 answers:

maria [59]3 years ago

8 0

Answer:

x^2-3x+36

Step-by-step explanation:

Sladkaya [172]3 years ago

5 0

Subtract the area of the inside rectangle from the outside one

THE ANSWER IS: x^2 - 3x + 36

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Find the slope of the line !

KatRina [158]

Answer:

Slope = 0

Step-by-step explanation:

Because there is no change, it's constant, so its equation is y = -1

3 0

4 years ago

Read 2 more answers

The vertices of a rectangle are located at (1, 2) (5, 0) (2, -6) and (-2, -4) what is the area of the rectangle

galina1969 [7]

Answer:

Option 2) 30 square units

Step-by-step explanation:

step 1

Plot the rectangle to better understand the problem

we have

A(1, 2),B(5, 0),C(2, -6) and D(-2, -4)

see that attached figure

step 2

Find the length of the sides

Remember that'

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

<em>Find the length side AB</em>

we have

A(1, 2),B(5, 0)

substitute in the formula

$d=\sqrt{(0-2)^{2}+(5-1)^{2}}$

$d=\sqrt{(-2)^{2}+(4)^{2}}$

$d_A_B=\sqrt{20}\ units$

<em>Find the length side BC</em>

we have

B(5, 0),C(2, -6)

substitute in the formula

$d=\sqrt{(-6-0)^{2}+(2-5)^{2}}$

$d=\sqrt{(-6)^{2}+(-3)^{2}}$

$d_B_C=\sqrt{45}\ units$

step 3

Find the area of rectangle

The area of rectangle is equal to

$A=LW$

we have

$L=BC=\sqrt{45}\ units$

$W=AB=\sqrt{20}\ units$

substitute

$A=(\sqrt{45})(\sqrt{20})=\sqrt{900}=30\ units^2$

7 0

3 years ago

PLEESE HELP ME I NEED IT PLEASE THIS IS ALL MY PIONTS −4 3/ 5 ÷1 1 /5

Vladimir [108]

let's firstly convert the mixed fractions to improper fractions and then divide.

$\bf \stackrel{mixed}{-4\frac{3}{5}}\implies \cfrac{-4\cdot 5+3}{5}\implies \stackrel{improper}{\cfrac{-23}{5}}~\hfill \stackrel{mixed}{1\frac{1}{5}}\implies \cfrac{1\cdot 5+1}{5}\implies \stackrel{improper}{\cfrac{6}{5}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{-23}{5}\div \cfrac{6}{5}\implies \cfrac{-23}{\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}\cdot \cfrac{\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}{6}\implies \cfrac{-23}{6}\implies -3\frac{5}{6}$